Đáp án:
b) \(\dfrac{{x - 1}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
x \ne 0\\
x + 5 \ne 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne 0\\
x \ne - 5
\end{array} \right.\\
\to x \ne \left\{ { - 5;0} \right\}\\
b)P = \dfrac{{{x^2} + 2x}}{{2\left( {x + 5} \right)}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2\left( {x - 5} \right)\left( {x + 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2\left( {{x^2} - 25} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 4x - 5}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 5x - x - 5}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {x + 5} \right) - \left( {x + 5} \right)}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{\left( {x + 5} \right)\left( {x - 1} \right)}}{{2\left( {x + 5} \right)}} = \dfrac{{x - 1}}{2}
\end{array}\)
